Denote the logarithm to base 10 with log and the largest integer smaller x with [x]. Assuming you have a large enough one, you can calculate B-A as follows:

We have[log(A)]=[79641170620168673833*log(2)]=23974381246463762440[log(B)]=[50247984153525417450*log(3)]=23974381246463762440These are the numbers of digits minus one of A and B. They are identical, so nothing to go on yet.

10^(log(A)-[log(A)])=0.5076025219062843177372568120345012380597232563753110^(log(B)-[log(B)])=0.50760252190628431774066783531935296206309427065819These are the mantissa of A and B with 50-digit or so accuracy. We can already tell B > A.

10^(log(B)-[log(B)])-10^(log(A)-[log(A)])=0.341102328485172400337101428288*10^(-20)This is the mantissa of b-a time * 10^(-20), i.e. 20 zero's to shift to the left to get a real mantissa.

[log(A)]-20=23974381246463762420This is the number of digits minus one of B-A. So we finally get

B-A=0.341102328485172400337101428288*10^23974381246463762420Not quite as big as A or B, but still quite big.

Another way to determine B > A without the need of an extremely large calculator (you'll still need one bigger than average), is to speculate how Frederico in his cunning ways has come up with the exponents of A and B. If we wanted A and B "as equal as possible", we were to approximately solve 2^x=3^y in integers x and y. Follows x*log(2)=y*log(3), follows x/y=log(3)/log(2). So we have to find a fraction x/y that approximates log(3)/log(2) as closely as possible. The prefered method for approximating numbers with fractions is to use continued fractions, as these are in a way best-approximations. After a few attempts you'll find that indeed 79641170620168673833 / 50247984153525417450 is the continued fraction of log(3)/log(2) terminated after 40 fractions. Because we know that continued fractions converge to the approximating number in an oscillating fashion (smaller, bigger, smaller, bigger,...) and since the first approximation is 1+1/(1+1) = 3/2, which is smaller than log(3)/log(2), the 40th approximation must be larger than log(3)/log(2). Going backwards it follows that B > A.